select n 基于排名列表的 python 列表中排名靠前的名称

Question

I have a sorted python list of ranks and a corresponding list of names. How do I get all the top 5 names in the order of the rank from the python list rank?

rank: [ 2, 1, 4, 1, 1, 3, 1]
names:  ['X', 'Y', 'Z', 'A', 'B', 'C', 'D']

desired output:

['X', 'Y', 'A', 'B', 'D']

I got this.. but does not limit to top 5

[ j for (i,j) in zip(rank, names) if i <= 2 ]

Answer 1

You can use sorted with zip :

In [244]: [x for _, x in sorted(zip(rank, names))][:5]
Out[244]: ['A', 'B', 'D', 'Y', 'X']

Answer 2

You need to sort twice to get the required result (Itemgtter is to extract the value at index 0)-

from operator import itemgetter
result = list(map(itemgetter(0),sorted(sorted(zip(names,rank),key=lambda x:x[1])[:5],key=lambda x:x[1],reverse=True)))
print(result ) # ['X', 'Y', 'A', 'B', 'D']